java java正则表达式从更大的字符串中排除特定的字符串
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java regex to exclude specific strings from a larger one
提问by nvrs
I have been banging my head against this for some time now:
I want to capture all [a-z]+[0-9]?character sequences excluding strings such as sin|cos|tanetc.
So having done my regex homework the following regex should work:
一段时间以来，我一直在反对这个问题：我想捕获[a-z]+[0-9]?除字符串等之外的所有字符序列sin|cos|tan。因此，在完成我的正则表达式作业后，以下正则表达式应该可以工作：
(?:(?!(sin|cos|tan)))\b[a-z]+[0-9]?
As you see I am using negative lookahead along with alternation - the \bafter the non-capturing group closing parenthesis is critical to avoid matching the inof sinetc. The regex makes sense and as a matter of fact I have tried it with RegexBuddy and Java as the target implementation and get the wanted result but it doesn't work using Java Matcher and Pattern objects!
Any thoughts?
如您所见，我正在使用负前瞻和交替 -\b在非捕获组右括号之后对于避免匹配inofsin等至关重要。正则表达式是有道理的，事实上我已经尝试过使用 RegexBuddy 和 Java 作为目标实现并获得想要的结果，但它不能使用 Java Matcher 和 Pattern 对象！有什么想法吗？
cheers
干杯
回答by bobince
The \bis in the wrong place. It would be looking for a word boundary that didn't have sin/cos/tan beforeit. But a boundary just afterany of those would have a letter at the end, so it would have to be an end-of-word boundary, which is can't be if the next character is a-z.
该\b是放错了地方。它要寻找那些没有正弦/余弦/棕褐色单词边界之前它。但是紧跟在其中任何一个之后的边界将在末尾有一个字母，因此它必须是词尾边界，如果下一个字符是 az，则不能是字尾边界。
Also, the negative lookahead would (if it worked) exclude strings like cost, which I'm not sure you want if you're just filtering out keywords.
此外，否定前瞻将（如果有效）排除像 那样的字符串cost，如果您只是过滤掉关键字，我不确定您是否想要。
I suggest:
我建议：
\b(?!sin\b|cos\b|tan\b)[a-z]+[0-9]?\b
Or, more simply, you could just match \b[a-z]+[0-9]?\band filter out the strings in the keyword list afterwards. You don't always have to do everything in regex.
或者，更简单地说，您可以在之后匹配\b[a-z]+[0-9]?\b并过滤关键字列表中的字符串。您不必总是在正则表达式中做所有事情。
回答by Tomalak
So you want [a-z]+[0-9]?(a sequence of at least one letter, optionally followed by a digit), unlessthat letter sequence resembles one of sincostan?
所以你想要[a-z]+[0-9]?（至少一个字母的序列，可选地后跟一个数字），除非该字母序列类似于sincostan?
\b(?!(sin|cos|tan)(?=\d|\b))[a-z]+\d?\b
results:
结果：
cos - no match cosy - full match cos1 - no match cosy1 - full match bla9 - full match bla99 - no match
回答by nvrs
i forgot to escape the \bfor java so \bshould be \\band it now works.
cheers
我忘了转义\bfor java 所以\b应该是\\b，它现在可以工作了。干杯
